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# S2 Poisson sum of 2 means problem - prob of 2 error types watch

1. Problem:

screwdrivers from trial production run inspected. Found that cellulose acetate handles are defective on 1% and the chrome steel blades are defective on 1.5% if the screwdrivers, the defects occurring independently.

(i) What is the probability that a sample of 80 contain more than 2 defective screwdrivers.

First bit I could answer ok.

H~Poisson(0.8)
B~Poisson(1.2)
T~Poisson(2) T = total, handle and blade

P(X > 2) = 0.3233 - this I answered ok.

(ii) what is the probability that a sample of 80 contains at least one screwdriver with both a defective handle and a defective blade?

Basically it is asking for the probability that at least one screwdriver in sample has both types of error.

My thinking was this is an AND question - so probability would be Prob1 multiplied by prob2. So I did like this:

H~Poisson(0.8)
B~Poisson(1.2)

So we want to find P(H >= 1) AND P(B >= 1)

P(H >= 1) = 0.5507
P(B >= 1) = 0.6988

Both: 0.5507 x 0.6988 = 0.3848

How do they get to 0.012???

How do you work out this sort of thing?
2. (Original post by acomber)
How do you work out this sort of thing?
Consider one screwdriver to start.

What's the probability that it has both defects?

So, what the probabilty that it doesn't have both defects?

Then, what's the probability that 80 don't have both defects?

Hence probabilty that at least one has both defects?

No Poisson required.
3. (Original post by ghostwalker)
Consider one screwdriver to start.

What's the probability that it has both defects?

So, what the probabilty that it doesn't have both defects?

Then, what's the probability that 80 don't have both defects?

Hence probabilty that at least one has both defects?

No Poisson required.
Consider one screwdriver to start.

What's the probability that it has both defects?

H~B(80, 0.01)
B~B(80, 0.015)

P(H=1) = 80C1 . 0.99^79 . 0.01 = 0.3616

P(B=1) = 80C1 . 0.985^79 . 0.015 = 0.3636

Sp probability that 1 screwdriver has both defects is:

0.3616 x 0.3636 = 0.1315

Is that correct so far?

So, what the probabilty that it doesn't have both defects?

P(1 screwdriver does NOT have both defects) = 1 - 0.1315

Is that correct?

Then, what's the probability that 80 don't have both defects?

** How do I do this step???

Hence probabilty that at least one has both defects?
4. (Original post by acomber)
Consider one screwdriver to start.

What's the probability that it has both defects?

H~B(80, 0.01)
B~B(80, 0.015)

P(H=1) = 80C1 . 0.99^79 . 0.01 = 0.3616

P(B=1) = 80C1 . 0.985^79 . 0.015 = 0.3636
Those are the probability that in a batch of 80, exactly one has a given defect.

1% of handles are defective, so probability that any given screwdriver has a defective handle is 0.01

Sp probability that 1 screwdriver has both defects is:

0.3616 x 0.3636 = 0.1315

Is that correct so far?
Yes, multiply the two, but you have the wrong figure.

So, what the probabilty that it doesn't have both defects?

P(1 screwdriver does NOT have both defects) = 1 - 0.1315

Is that correct?

Then, what's the probability that 80 don't have both defects?

** How do I do this step???
If probability that one doesn't have the defect is p, then the probability that 80 don't is p^80.

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